# Launching ICBMs into the Sun

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So I was hitting random on XKCD and this one came up: https://xkcd.com/1626/

I've seen discussions elsewhere about the delta-V needed to send something into the sun, and while I have no idea what the delta-V of the typical ICBM is, I doubt they even have enough to reach LEO since they're a weapon that's intended to hit a target somewhere else on the Earth. But just as a thought experiment, what if the XKCD scenario happened? Say you launched them retrograde to Earth's orbit. Would they just just rain down on Earth again and kill everyone? How much delta-V would be needed for one to escape Earth and not come back? (Not necessarily hitting the Sun, just ending up in a Solar orbit that never intersects Earth again).

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oh, what you ask is quite easy: it is the escape velocity from Earth's gravity well, which is 11.2 Km/s at its surface. Of course you need more than 11.2 Km/s Delta-V to compensate for atmospheric drag during the ascent (can't give sound numbers here, depends on TWR and aerodynamics of the rocket).

Really, to hit the Sun then would be needed to drop all the orbital speed the rocket still has by being co-orbital with Earth, that is about 30 Km/s on average. Possibly by using other bodies to craft a few gravity assists instead of using fuel.

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Anyone know how much delta-V an ICBM typically has? How much of a booster (MOAR BOOSTAS!!) would it take to get one on a solar orbit that never intersects Earth?

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They will reach 5000-6000 km above the Earth and fall down because thei delta-V (7-7.5 km/s) is much lower than the escape velocity.

But sentient computers would make them fall in one place and then bury under ground with one explosion under a mountain next to them.

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43 minutes ago, capran said:

Anyone know how much delta-V an ICBM typically has? How much of a booster (MOAR BOOSTAS!!) would it take to get one on a solar orbit that never intersects Earth?

dV values do not seem to be published, it would seem prudent for exact figures to be classified. But it can be estimated from data that is published, and also from observing their performance. Some good science here:

As a rough guide, many ICBMs are capable of putting something into low orbit, which gives you an idea of their dV capabilities. It aslo obviously depends heavily on the payload, which is not necessarily fixed per missile.

54 minutes ago, diomedea said:

oh, what you ask is quite easy: it is the escape velocity from Earth's gravity well, which is 11.2 Km/s at its surface. Of course you need more than 11.2 Km/s Delta-V to compensate for atmospheric drag during the ascent (can't give sound numbers here, depends on TWR and aerodynamics of the rocket).

Really, to hit the Sun then would be needed to drop all the orbital speed the rocket still has by being co-orbital with Earth, that is about 30 Km/s on average. Possibly by using other bodies to craft a few gravity assists instead of using fuel.

Is it not true that you can do it with significantly less dV than this if you do...what is it called?...a bi-elliptic transfer? Raise solar apoapsis first, then when you reach apoapsis, burn to lower periapsis into the sun?

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1 minute ago, p1t1o said:

Is it not true that you can do it with significantly less dV than this if you do...what is it called?...a bi-elliptic transfer? Raise solar apoapsis first, then when you reach apoapsis, burn to lower periapsis into the sun?

Not being a rocket scientist but being a KSPer with nearly 1800 hours, I think this checks out! ;P Have to try it in game.

But that does pre-suppose that the rocket on the ICBM is capable of being restarted. Also, what does an ICBM use for steering? Something like SAS or RCS? Thrust-vectoring? Would one even be capable of changing direction when it gets to apoapsis?

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@p1t1o: bi-elliptic transfers work great if the result you want is to attain a specific orbit (so, matching Ap and Pe with what desired, one first aims to meet the Ap and therefore enters one specific elliptic orbit, than when at the Ap burns to meet the Pe, resulting in a different elliptic orbit). For hitting the Sun, what is required is to decrease Pe to a value lower than the Sun radius, no need at all to change the Ap.

Earth's orbit around Sun is slightly elliptic, so a very small change in orbital speed exists; to make that maneuver, would be better to be at the Aphelion (so our speed is already lower, and will require less to drop it for lowering the Pe).

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They just push each other... Launch on ICBM into the next... knock them up into the sun as a kind of ad-hoc space fountain.

Ps, KSP does not do multibody gravity, so I assume would give no benefit from trying to gain DV/lessen DV requirement by launching "towards" the suns pull...

Edited by Technical Ben
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12 minutes ago, capran said:

Not being a rocket scientist but being a KSPer with nearly 1800 hours, I think this checks out! ;P Have to try it in game.

But that does pre-suppose that the rocket on the ICBM is capable of being restarted. Also, what does an ICBM use for steering? Something like SAS or RCS? Thrust-vectoring? Would one even be capable of changing direction when it gets to apoapsis?

I dont think they can be re-started, but the last stage has RCS for attitude control and a little extra dV (it uses this to place multiple warheads onto different targets).

**edit**

@p1t1o: bi-elliptic transfers work great if the result you want is to attain a specific orbit (so, matching Ap and Pe with what desired, one first aims to meet the Ap and therefore enters one specific elliptic orbit, than when at the Ap burns to meet the Pe, resulting in a different elliptic orbit). For hitting the Sun, what is required is to decrease Pe to a value lower than the Sun radius, no need at all to change the Ap.

Earth's orbit around Sun is slightly elliptic, so a very small change in orbital speed exists; to make that maneuver, would be better to be at the Aphelion (so our speed is already lower, and will require less to drop it for lowering the Pe).

Unless I am missing something...and I might be!  .....

It sounds like you are describing a standard Hohman transfer - what I am alluding to is first burning to raise your (solar) Ap waaaaay above what you need it to be, then when you reach this new, higher Ap, you can alter your Pe for super-cheap. The savings on changing the Pe will offset the extra burn to initially raise the Ap. I think this even holds when you want to re-circularise as well, but in this case we are plunging into the sun

Edited by p1t1o
avoid double post
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@p1t1o: well, I'm not about to rule out definitely what you suggested, I saw similar ideas in the past and I'm sure those work in some cases. But while achieving a higher Ap definitely helps with other maneuvers, I'm not so sure it does for just lowering Pe when starting from a an almost perfectly circular orbit. May have to compute how it goes, my initial feeling is the energy added for raising Ap just balances out with the (less) energy required for then lowering Pe.

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1 minute ago, diomedea said:

@p1t1o: well, I'm not about to rule out definitely what you suggested, I saw similar ideas in the past and I'm sure those work in some cases. But while achieving a higher Ap definitely helps with other maneuvers, I'm not so sure it does for just lowering Pe when starting from a an almost perfectly circular orbit. May have to compute how it goes, my initial feeling is the energy added for raising Ap just balances out with the (less) energy required for then lowering Pe.

Unless Im way off, it should be possible to get significant savings, fortunately we both come equipped with a simulator perfectly suited to this problem!

I know, I know. Its a huge pain, but I guess we'll just have to play some more KSP...

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I'll see your xkcd and raise you Scott Manley:

[too long; didn't watch] Matching Earth's orbital velocity in delta-v is *expensive*.  For such large delta-v changes, your best bet is a Jupiter slingshot.  Even so, I shouldn't have to point out that getting an extra ~6500m/s (beyond Earth orbit, which is more than an ICBM needs) out of an ICBM is pretty much impossible.

Edited by wumpus
s/an/and
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@p1t1o: and you are right! Didn't test in KSP (though you're welcome to do if you like), but computed the maneuver. Sorry if math below is less than clear.

Vpe = SQRT(G*M*(2/Rpe - 1/SMA)) gives the speed to have at periapsis (known gravitational constant G and mainbody mass M, orbit height at periapsis Rpe and orbit semimajor axis SMA), this allows to know how much Delta V to put in the first burn to have the apoapsis raise to Rap = 2*SMA - Rpe.

Then, speed at apoapsis Vap = SQRT(G*M*(2/Rap - 1/SMA)). That is exactly the amount we have to deplete with the second burn.

E.g. in KSP: Kerbin's orbit SMA=Ap=Pe is 1.35998E10, Sun mass = 1.75655E28, so Kerbin moves at 9.284 Km/s (that would be the same speed of an ICBM launched from Kerbin with just the escape velocity to enter a Sun orbit). Then the ICBM burns prograde in Sun orbit, to raise the aphelion (Rap) to, say, 3.35998E10 (2E10 meters higher): new Vpe = 11.078 Km/s (so, Delta V = 1794 m/s). At the aphelion the speed Vap = 4.484 Km/s that is the amount to deplete. Therefore the total Delta V required with the maneuver you suggested is 1794 + 4484 = 6278 m/s that is well lower than the 9284 m/s speed in the original circular orbit.

Anyway, I'd still go with some gravity assist (e.g., a multiple pass by Eve, Moho, Eve to keep reducing orbital energy until needed; if done right won't cost more than doing the initial Eve intercept, that's lower than the bi-elliptic burns).

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An ICBM might make orbit with a lightened payload - after all plenty of orbital launchers have been based on missiles. But I strongly doubt it could escape Earth. Randall knows this, that's why he mentions booster rockets.

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I vaguely recall some back-of-the-envelope calculations that show that it's cheaper (DV-wise at least) to just launch something out of the solar system than it is to launch it into the sun. I might be wrong though.

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Didn't read thread, but if you need solar system delta-v I have a nice pic).

About 20km/s of dV is what you're looking at for a Sun dive from LEO. An escape trajectory is around 10km/s. Taking gravity assists can cut that down to a handful km/s though (basically what you need to get to Jupiter and then head to wherever you want).

Definitely doable with reasonable rockets (most rockets are ICBMs without a non-explody thing on top). Dawn can pull out 10km/s of dV and weighs 1.2t which should be about the mass of your average warhead. Put more fuel, launch it on top whatever launcher you want and you're good to go.

Don't really see the point of it though, since the Sun outputs a few dozen petatons (billions of megatons) of TNT per second, but yeah, you can do it..

Edited by Gaarst
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Escape velocity plus Earth's orbital velocity. Or... lots. 11kps or so plus 30 kps.

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Take this as a starting point. According to the link, its mass is 1700 kg. You need 11.2-7.9 =  3.3 km/s more just to leave the Earth orbit.
Now you can calculate what mass can reach the hyperbolic orbit (or to hit the Moon) if replace this head with subrocket with total mass 1700 kg.

Don't forget that reaching 11.2 you just begin to revolve round the Sun near the Earth. It requires ~10-15 km/s more to reach something else.

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18 hours ago, capran said:

Would they just just rain down on Earth again and kill everyone?

Tangent: they'd likely just rain down on Eath again and make holes in the ground... nobody is stupid enough to make a nuclear warhead that detonates on impact (without a ridiculously complex arming sequence anyway).
On topic: An ICBM (with warhead, which is the point) won't make orbit, let alone escape velocity. It's gonna come down somewhere.

Edited by steve_v
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The heat shields of ICBMs are designed to survive reentry from a shallow suborbital trajectory.  They would burn up on reentry when being launched straight up.

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16 hours ago, Kerbart said:

I vaguely recall some back-of-the-envelope calculations that show that it's cheaper (DV-wise at least) to just launch something out of the solar system than it is to launch it into the sun. I might be wrong though.

If  Vo~= Ve/(2**.5) ["2**.5" = "square root of 2" in FORTRAN and Python] (an equation that keeps popping up in my googling, note that this is absolute minimum orbital velocity: screaming over a perfectly spherical [cow with no atmosphere] at zero height).

Then you need only add 40% of your [i.e. Earth's] orbital velocity to escape the Solar System.  To plunge into the Sun, you will need to kill 100% (roughly) of your orbital velocity.  I might be reading my subway map wrong, but there is a reason why Mohoo takes the most delta-v to get to.

Edited by wumpus
cut and pasted formula didn't copy
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you could boost the deltaV by flying in a close formation with one icbm trailing behind and triggering the warhead... primitive orion drive anyone?^^

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1 hour ago, hms_warrior said:

you could boost the deltaV by flying in a close formation with one icbm trailing behind and triggering the warhead... primitive orion drive anyone?^^

Great idea, but I don't really think ICBM engines can be restarted.  You'd need plenty of adjustment burns and all other tricks on you journey to get enough planetary kicks to get to the Sun.

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1 minute ago, wumpus said:

but I don't really think ICBM engines can be restarted.

Who says about "engine restart"? The suggestion was to trigger a warhead to push another rocket above. An improvised Orion.

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2 hours ago, hms_warrior said:

you could boost the deltaV by flying in a close formation with one icbm trailing behind and triggering the warhead... primitive orion drive anyone?^^

The amount of modifications you would need to do to get this to work is enough to warrant calling it "building an entirely new spacecraft". Modifications would be needed even to the warheads themselves.

If you just detonate a nuke behind an ICBM bus, the propellant is going to be...bits of the ICBM bus.

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